Lesson 7.2 Solving a System of Linear Equations ...hrsbstaff.ednet.ns.ca/mandrecyk/Math...

Pearson Chapter 7 Foundations and Pre-calculus Mathematics 10 Systems of Linear Equations

Lesson 7.2 Copyright 2011 Pearson Canada Inc. 1

Lesson 7.2 Solving a System of Linear Equations Graphically Exercises (pages 409-410)

A 3. The solution is the coordinates of the point of intersection of the two lines on each graph.

a) The solution is: x = –4 and y = 2

b) The solution is: x = 2 and y = 3

c) The solution is: x = 1 and y = –3

d) The solution is: x = –2 and y = –1

B 4. a) The solution appears to be: x = 9 and y = –2

Substitute these values for x and y into each equation to check.

2x + 3y = 12 x – y = 11

L.S. = 2x + 3y L.S. = x – y

= 2(9) + 3(–2) = 9 – (–2)

= 18 – 6 = 9 + 2

= 12 = 11

= R.S. = R.S.

For each equation, the left side is equal to the right side.

So, x = 9 and y = –2 is a solution of the linear system. This solution is exact.

b) The solution appears to be: x = –13

4 and y = 2

3

4


–3x + y = 31

4 3x – 4y = –16

L.S. = –3x + y L.S. = 3x – 4y

= –33

14

+ 23

4 = 3

31

4

– 43

24

= (–3)7

4

+ 11

4 = 3

7

4

– 411

4

= 21

4 +

11

4 =

21

4 –

44

4

= 32

4 =

65

4 , or –16

1

4

For each equation, the left and right sides are not equal, but they are close in value.

So, x = –13

4 and y = 2

3

4 is an approximate solution of the linear system.

5. a) i) x + y = 7

3x + 4y = 24

Determine the x- and y-intercepts of the graph of each equation.

For equation :

x + y = 7



When x = 0, When y = 0,

0 + y = 7 x + 0 = 7

y = 7 x = 7

On a grid, mark a point at 7 on each axis, then draw a line through the points.

For equation :

3x + 4y = 24


3(0) + 4y = 24 3x + 4(0) = 24

4y = 24 3x = 24

y = 6 x = 8

On the grid, mark a point at 6 on the y-axis and mark a point at 8 on the x-axis, then

draw a line through the points.

The point of intersection appears to be: (4, 3)

Verify the solution.

Substitute x = 4 and y = 3 into each equation.

x + y = 7 3x + 4y = 24

L.S. = x + y L.S. = 3x + 4y

= 4 + 3 = 3(4) + 4(3)

= 7 = 12 + 12

= R.S. = 24

= R.S.

Since the left side is equal to the right side in each equation, then x = 4 and y = 3 is

the solution of the linear system.

ii) x – y = –1

3x + 2y = 12


For equation :

x – y = –1


0 – y = –1 x – 0 = –1

y = 1 x = –1

On a grid, mark a point at 1 on the y-axis and mark a point at –1 on the x-axis, then


For equation :

3x + 2y = 12


3(0) + 2y = 12 3x + 2(0) = 12

2y = 12 3x = 12

y = 6 x = 4



On the grid, mark a point at 6 on the y-axis and mark a point at 4 on the x-axis, then


The point of intersection appears to be: (2, 3)


Substitute x = 2 and y = 3 into each equation.

x – y = –1 3x + 2y = 12

L.S. = x – y L.S. = 3x + 2y

= 2 – 3 = 3(2) + 2(3)

= –1 = 6 + 6

= R.S. = 12

= R.S.

Since the left side is equal to the right side in each equation, then x = 2 and y = 3 is


iii) 5x + 4y = 10

5x + 6y = 0


For equation :

5x + 4y = 10


5(0) + 4y = 10 5x + 4(0) = 10

4y = 10 5x = 10

y = 10

4, or 2

1

2 x = 2

Since the y-intercept is not where grid lines intersect, determine the coordinates of

another point on the line.

Substitute: x = 6

5(6) + 4y = 10

30 + 4y = 10

4y = –20

y = –5

On a grid, mark a point at (6, –5) and mark a point at 2 on the x-axis, then draw a line

through the points.

For equation :

5x + 6y = 0


5(0) + 6y = 0 5x + 6(0) = 0

6y = 0 5x = 0

y = 0 x = 0

Since the x-and y-intercepts are equal, determine the coordinates of another point on

the line.



Choose a value for x that is a multiple of 6, so the corresponding y-value is an

integer.

Substitute: x = 6

5(6) + 6y = 0

30 + 6y = 0

6y = –30

y = –5

On the grid, mark a point at (6, –5) and mark a point at the origin, then draw a line

through the points.

The point of intersection appears to be: (6, –5)

Since the coordinates of this point were determined using each equation, these

coordinates satisfy each equation and x = 6 and y = –5 is the solution of the linear

system.

iv) x + 2y = –1

2x + y = –5


For equation :

x + 2y = –1


0 + 2y = –1 x + 2(0) = –1

2y = –1 x = –1

y = 1

2

Since the y-intercept is not where grid lines intersect, determine the coordinates of


Substitute: x = 1

1 + 2y = –1

2y = –2

y = –1

On a grid, mark a point at (1, –1) and mark a point at –1 on the x-axis, then draw a

line through the points.

For equation :

2x + y = –5


2(0) + y = –5 2x + 0 = –5

y = –5 2x = –5

x = –21

2

Since the x-intercept is not where grid lines intersect, determine the coordinates of




Substitute: x = –1

2(–1) + y = –5

–2 + y = –5

y = –3

On the grid, mark a point at (–1, –3) and mark a point at –5 on the y-axis, then draw a


The point of intersection appears to be: (–3, 1)


Substitute x = –3 and y = 1 into each equation.

x + 2y = –1 2x + y = –5

L.S. = –3 + 2(1) L.S. = 2x + y

= –3 + 2 = 2(–3) + 1

= –1 = –6 + 1

= R.S. = –5

= R.S.

Since the left side is equal to the right side in each equation, then x = –3 and y = 1 is


b) For the linear system in part a, iv:

x + 2y = –1

2x + y = –5

The point of intersection of the graphs is the point with coordinates (–3, 1).

This is the solution of the linear system. This point lies on both lines of the system, and

its coordinates satisfy both equations of the system.

6. The linear system is:

3x – y = 1149

–x + 2y = 142

Substitute the solution (500, 300) to check whether it is exact or approximate.

3x – y = 1149 –x + 2y = 142

L.S. = 3x – y L.S. = –x + 2y

= 3(500) – 300 = –500 + 2(300)

= 1500 – 300 = –500 + 600

= 1200 = 100

For each equation, the left and right sides are not equal, but are close in value.

So, the solution x = 500 and y = 300 is an approximate solution.

7. a) 2x + 4y = –1

3x – y = 9

Graph each equation.

For equation :

2x + 4y = –1



Since the coefficients of x and y are even and the constant term is odd, there are no

integer coordinates that satisfy the equation.

Determine the x- and y-intercepts, then graph the equation on a grid with a larger scale.

2x + 4y = –1


2(0) + 4y = –1 2x + 4(0) = –1

4y = –1 2x = –1

y = 1

4 x =

1

2

On a grid, use a scale of 4 squares represents 1 unit, mark a point at 1

4 on the y-axis

and mark a point at 1

2 on the x-axis, then draw a line through the points.

For equation :

3x – y = 9


3(0) – y = 9 3x – 0 = 9

–y = 9 3x = 9

y = –9 x = 3

For a y-intercept of –9, the y-axis would be very long on a graph with a scale of 4 squares

to 1 unit, so choose a different point to plot.

Substitute x = 2 in 3x – y = 9.

3(2) – y = 9

6 – y = 9

y = –3

On the grid, mark a point at (2, –3) and mark a point at 3 on the x-axis.

Draw a line through the points.

The solution appears to be: x = 21

2 and y =

11

2

Write these numbers as decimals: x = 2.5 and y = –1.5




2x + 4y = –1 3x – y = 9

L.S. = 2x + 4y L.S. = 3x – y

= 2(2.5) + 4(–1.5) = 3(2.5) – (–1.5)

= 5 – 6 = 7.5 + 1.5

= –1 = 9

= R.S. = R.S.

Since the left side is equal to the right side for each equation, the solution of the linear

system is x = 2.5 and y = –1.5.

b) 5x + 5y = 17

x – y = –1


For equation :

5x + 5y = 17

Since the coefficients of x and y are even and the constant term is odd, there are no

integer coordinates that satisfy the equation.

Determine the x-and y-intercepts, then graph the equation on a grid with a larger scale.

5x + 5y = 17


5(0) + 5y = 17 5x + 5(0) = 17

5y = 17 5x = 17

y = 17

5, or 3

2

5 x =

17

5, or 3

2

5


5 on the y-axis and

mark a point at 32

5 on the x-axis, then draw a line through the points.

For equation :

x – y = –1


0 – y = –1 x – 0 = –1

y = 1 x = –1

On the grid, mark a point at 1 on the y-axis and mark a point at –1 on the x-axis.





5 and y =

12

5

Write these numbers as decimals: x = 1.2 and y = 2.2


5x + 5y = 17 x – y = –1

L.S. = 5x + 5y L.S. = x – y

= 5(1.2) + 5(2.2) = 1.2 – 2.2

= 6 + 11 = –1

= 17 = R.S.

= R.S.


system is: x = 1.2 and y = 2.2

c) x + y = 23

4

x – y = 3

4


For equation :

x + y = 23

4

Since the constant term is a rational number, there are no integer coordinates that satisfy

the equation.

Determine the x- and y-intercepts, then graph the equation on a grid with a larger scale.

x + y = 23

4


0 + y = 23

4 x + 0 =

23

4

y = 23

4, or

35

4 x =

23

4, or

35

4




54

on the y-axis and

mark a point at 3

54

on the x-axis, then draw a line through the points.

For equation :

x – y = 3

4


0 – y = 3

4 x – 0 =

3

4

y = –3

4 x =

3

4

On the grid, mark a point at –3

4 on the y-axis and mark a point at

3

4 on the x-axis.



4 and y =

12

2

Write these numbers as decimals: x = 3.25 and y = 2.5

Substitute these values for x and y into each equation to check. Write the rational

numbers in the equations as decimals.

x – y = 3

4 x + y =

23

4

x – y = 0.75 x + y = 5.75

L.S. = x – y L.S. = 3.25 + 2.5



= 3.25 – 2.5 = 5.75

= 0.75 = R.S.

= R.S.


system is x = 3.25 and y = 2.5.

d) 3x + y = 6

x + y = 4

3


For equation :

3x + y = 6

Determine the x- and y-intercepts.

3x + y = 6


3(0) + y = 6 3x + 0 = 6

y = 6 3x = 6

x = 2

On a grid, mark a point at 6 on the y-axis and mark a point at 2 on the x-axis, then draw a


For equation :

x + y = 4

3


0 + y = 4

3 x – 0 =

4

3

y = 4

3 x =

4

3

On the grid, use a scale of 3 squares to 1 unit. Mark a point at x + y = 4

3 on the y-axis

and mark a point at 4

3 on the x-axis.





3 and y = –5


x + y = 4

3 3x + y = 6

L.S. = x + y L.S. = 3x + y

= 32

3 – 5 = 3

23

3

– 5

= 11

3 –

15

3 = (3)

11

3

– 5

= 4

3 =

33

3 – 5

= R.S. = 11 – 5

= 6

= R.S.


system is x = 32

3 and y = –5.



8. a) C = 175 + 0.10n

C = 250 + 0.07n

For each equation, determine the C-intercept and the coordinates of another point on the

line.

For equation :

C = 175 + 0.10n

Substitute: n = 0

C = 175

Substitute: n = 1000

C = 175 + 0.10 × 1000

C = 175 + 100

C = 275

On a grid, use a scale of 2 squares to 100 units on the C-axis, and a scale of 2 squares

to 500 units on the n-axis. Mark a point at 175 on the C-axis and mark a point at

(1000, 275). Draw a line through the points. The data are discrete, but the scale is so

small that if a point were plotted for each brochure, these points would form a line.

For equation :

C = 250 + 0.07n

Substitute: n = 0

C = 250

Substitute: n = 1000

C = 250 + 0.07 × 1000

C = 250 + 70

C = 320

On the grid, mark a point at 250 on the C-axis and mark a point at (1000, 320). Draw a


b) i) From the graph, the cost will be the same at both companies at the point of

intersection of the lines; that is, when 2500 brochures are printed for $425.

Check that this solution satisfies both equations.

Substitute C = 425 and n = 2500 in each equation.

For equation :

C = 175 + 0.10n

L.S. = C R.S. = 175 + 0.10n

= 425 = 175 + 0.10(2500)

= 175 + 250

= 425



For equation :

C = 250 + 0.07n

L.S. = C R.S. = 250 + 0.07n

= 425 = 250 + 0.07(2500)

= 250 + 175

= 425

Since the left side is equal to the right side for each equation, the solution is correct.

ii) The equation that represents Company A is C = 175 + 0.10n. It is cheaper to use

Company A when its line is beneath the line for Company B; that is, for any number

of brochures less that 2500.

9. a) P = 700 + 0.03s

P = 1000 + 0.02s

For each equation, determine the P-intercept and the coordinates of another point on the

line.

For equation :

P = 700 + 0.03s

Substitute: s = 0

P = 700

Substitute: s = 10 000

P = 700 + 0.03 × 10 000

P = 700 + 300

P = 1000

On a grid, use a scale of 1 square to 200 units on the P-axis, and a scale of 2 squares to

10 000 units on the s-axis. Mark a point at 700 on the P-axis and mark a point at (10 000,

1000). Draw a line through the points.

For equation :

P = 1000 + 0.02s

Substitute: s = 0

P = 1000

Substitute: s = 10 000

P = 1000 + 0.02 × 10 000

P = 1000 + 200

P = 1200

On the grid, mark a point at 1000 on the P-axis and mark a point at (1000, 1200). Draw a




b) i) From the graph, a clerk will receive the same salary with both plans at the point of

intersection of the lines; that is, when $30 000 in sales earns a salary of $1600.


Substitute P = 1600 and s = 30 000 in each equation.

For equation :

P = 700 + 0.03s

L.S. = P R.S. = 700 + 0.03s

= 1600 = 700 + 0.03(30 000)

= 700 + 900

= 1600

For equation :

P = 1000 + 0.02s

L.S. = P R.S. = 1000 + 0.02s

= 1600 = 1000 + 0.02(30 000)

= 1000 + 600

= 1600


ii) The equation that represents Plan B is P = 1000 + 0.02s. A person on Plan B is

earning more than he or she would on Plan A when the line for Plan B is above the

line for Plan A; that is, for any value of sales up to $30 000.

10. Let the area of the forested part be represented by f hectares.

Let the unforested area be represented by r hectares.

The area of Stanley park is 391 hectares. This is the sum of f hectares and r hectares.

So, one equation is: f + r = 391

The forested area is 141 hectares more than the unforested area.

So, another equation is: f = r + 141

A linear system is:

f + r = 391

f = r + 141

Graph the equations.

Determine the f- and r-intercepts.

For equation :

f + r = 391

Substitute: r = 0 Substitute: f = 0

f + 0 = 391 0 + r = 391

f = 391 r = 391

On a grid, plot r as a function of f; that is, plot r on the vertical axis and f on the horizontal

axis. Use a scale of 1 square to 50 units on each axis. Mark a point at 391 on each axis Draw

a line through the points. Since the point at 391 is an approximate position, the graph is

approximate.

For equation :

f = r + 141

Substitute: r = 0 Substitute: f = 0

f = 0 + 141 0 = r + 141

f = 141 r = –141

A negative value of r makes no sense when we are considering areas. This value of r is not in

the range. Substitute a different value of f, such as f = 200.

200 = r + 141

r = 59

On the grid, mark a point at 141 on the f-axis and mark a point at (200, 59).




From the graph, the coordinates of the point of intersection indicate the two areas.

The f-coordinate is approximately 265 and the r-coordinate is approximately 125.

So, the forested area is about 265 hectares and the rest of the park is about 125 hectares.

Use the given information to check this solution.

The sum of the areas, in hectares, is: 265 + 125 = 390; this is close to the given sum of 391.

The difference of the areas is: 265 – 125 = 140; this is close to the given difference of 141.

The solution is approximate.

11. Let the number of wins be represented by w.

Let the number of overtime losses be represented by l.

The team had a total of 107 points.

The team gets 2 points for a win and 1 point for an overtime loss.

So, one equation is: 2w + l = 107

The team had 43 more wins than overtime losses.

So, another equation is: w – l = 43

A linear system is:

2w + l = 107

w – l = 43


Determine the w- and l-intercepts.

For equation :

2w + l = 107

Substitute: l = 0 Substitute: w = 0

2w + 0 = 107 2(0) + l = 107

2w = 107 l = 107

w = 53.5

Since the w-intercept is not a whole number, substitute a different value of l, and determine

the corresponding w-coordinate.

Substitute: l = 17

2w + 17 = 107

2w = 90

w = 45

On a grid, plot l as a function of w; that is, plot l on the vertical axis and w on the horizontal

axis. Use a scale of 1 square to 10 units on each axis. Mark a point at (45, 17) and mark a

point at 107 on the l-axis. Since the data are discrete, the graph is a set of points that lie on a

line through the plotted points. Use a straightedge to mark a few points along the line at

whole number values of w and l.



For equation :

w – l = 43

Substitute: l = 0 Substitute: w = 0

w – 0 = 43 0 – l = 43

w = 43 l = –43

A negative value of l makes no sense when we are considering numbers of wins and overtime

losses. This value of l is not in the range. Substitute a different value of w, such as w = 80.

80 – l = 43

l = 37

On the grid, mark a point at 43 on the w-axis and mark a point at (80, 37). Since the data are

discrete, the graph is a set of points that lie on a line through the plotted points. Use a

straightedge to mark a few points along the line at whole number values of w and l.

From the graph, the coordinates of the point of intersection are (50, 7).

The w-coordinate is approximately 50 and the l-coordinate is approximately 7.

So, the number of wins is about 50 and the number of overtime losses is about 7.


The number of points is: 2(50) + 7 = 107; this is equal to the given number of points.

The difference in wins and overtime losses is: 50 – 7 = 43; this is equal to the given

difference.

The solution is correct and the numbers are exact.

12. Let the number of $5 gift cards sold be represented by f.

Let the number of $10 gift cards be represented by t.

The class raised $800.

So, one equation is: 5f + 10t = 800

The class sold a total of 115 gift cards.

So, another equation is: f + t = 115

A linear system is:

5f + 10t = 800

f + t = 115


Determine the f- and t-intercepts.

For equation :

5f + 10t = 800

Substitute: t = 0 Substitute: f = 0

5f + 10(0) = 800 5(0) + 10t = 800

5f = 800 10t = 800

f = 160 t = 80

On a grid, plot t as a function of f; that is, plot t on the vertical axis and f on the horizontal



axis. Use a scale of 1 square to 10 units on each axis. Mark a point at 160 on the f-axis and

mark a point at 80 on the t-axis. Since the data are discrete, the graph is a set of points that lie

on a line through the plotted points. Use a straightedge to mark a few points along the line at

whole number values of f and t.

For equation :

f + t = 115

Substitute: t = 0 Substitute: f = 0

f + 0 = 115 0 + t = 115

f = 115 t = 115

On the grid, mark a point at 115 on each axis. Since the data are discrete, the graph is a set of

points that lie on a line through the plotted points. Use a straightedge to mark a few points

along the line at whole number values of f and t.

From the graph, the coordinates of the point of intersection appear to be (70, 45).

The f-coordinate is approximately 70 and the t-coordinate is approximately 45.

So, the number of $5 gift cards sold is about 70 and the number of $10 gift cards sold is about

45.


The amount of money raised, in dollars, is: 5(70) + 10(45) = 350 + 450, or 800; this is equal

to the given amount.

The total number of gift cards sold is: 70 + 45 = 115; this is equal to the total number of cards

sold.


13. Let the number of student tickets sold be represented by s.

Let the number of adult tickets sold be represented by a.

The total admission fee was $152.

So, one equation is: 4.80s + 8a = 152

There were 13 more students than adults.

So, another equation is: s – a = 13

A linear system is:

4.80s + 8a = 152

s – a = 13


Determine the s- and a-intercepts.

For equation :

4.80s + 8a = 152



Substitute: a = 0 Substitute: s = 0

4.80s + 8(0) = 152 4.80(0) + 8a = 152

4.80s = 152 8a = 152

s = 31.6 a = 19

Since the s-intercept is not a whole number, substitute a different value for a, then solve for s.

Substitute: a = 10

4.80s + 8(10) = 152

4.80s + 80 = 152

4.80s = 72

s = 15

On a grid, plot a as a function of s; that is, plot a on the vertical axis and s on the horizontal

axis. Use a scale of 1 square to 2 units on each axis. Mark a point at 19 on the a-axis and

mark a point at (15, 10). Since the data are discrete, the graph is a set of points that lie on a

line through the plotted points. Use a straightedge to mark a few points along the line at

whole number values of a and s.

For equation :

s – a = 13

Substitute: a = 0 Substitute: s = 0

s – 0 = 13 0 – a = 13

s = 13 a = –13

A negative value of a makes no sense when we are considering numbers of adults. This value

of a is not in the range. Substitute a different value of s, such as s = 30.

30 – a = 13

a = 17

On the grid, mark a point at 13 on the s-axis and mark a point at (30, 17). Since the data are

discrete, the graph is a set of points that lie on a line through the plotted points. Use a

straightedge to mark a few points along the line at whole number values of s and a.


The s-coordinate is approximately 20 and the a-coordinate is approximately 7.

So, the number of student tickets sold is about 20 and the number of adult tickets sold is

about 7.


The total admission fee, in dollars, is: 4.8(20) + 7(8) = 96 + 56, or 152; this is equal to the

given admission fee.

The difference in the numbers of students and adults is: 20 – 7 = 13; this is equal to the given

difference.




14. a) Let the mass of the box be represented by b grams.

Let the mass of a golf ball be represented by g grams.

The box and 36 golf balls have a mass of 1806 g.

So, one equation is: b + 36g = 1806

When 12 balls are removed, 24 balls remain, so the mass of the box and 24 balls is

1254 g.

b + 24g = 1254

A linear system is:

b + 36g = 1806

b + 24g = 1254

b) To graph the equations, determine the b- and g-intercepts.

For equation :

b + 36g = 1806

Substitute: g = 0 Substitute: b = 0

b + 36(0) = 1806 (0) + 36g = 1806

b = 1806 36g = 1806

g = 50.16

The g-intercept is not a whole number, and it would be difficult to determine a whole

number value, so the graph and solution will be approximate.

On a grid, plot g as a function of b; that is, plot g on the vertical axis and b on the

horizontal axis. Use a scale of 1 square to 100 units on the b-axis and 1 square to 5 units

on the g-axis. Mark a point at 50 on the g-axis and mark a point at (1806). Since the data

are discrete, the graph is a set of points that lie on a line through the plotted points. Use a

straightedge to mark a few points along the line at whole number values of g.

For equation :

b + 24g = 1254

Substitute: g = 0 Substitute: b = 0

b + 24(0) = 1254 0 + 24g = 1254

b = 1254 24g = 1254

g = 52.25

As with equation , the g-intercept is not a whole number, so the graph and solution will

be approximate.

On the grid, mark a point at 1254 on the b-axis and mark a point at 52 on the g-axis.

Since the data are discrete, the graph is a set of points that lie on a line through the plotted

points. Use a straightedge to mark a few points along the line at whole number values of

g.




The b-coordinate is approximately 150 and the g-coordinate is approximately 45.

So, the mass of the box is about 150 g and the mass of a golf ball is about 45 g.


The mass of the box and 36 balls, in grams, is: 150 + 36(45) = 1770; this is close to the

given mass of 1806 g.

The mass of the box and 24 balls, in grams, is: 150 + 24(45) = 1230; this is close to the

given mass of 1254 g.

The mass of the box is about 150 g and the mass of one golf ball is about 45 g.

c) It was difficult to determine the exact solution because I could not plot the intercepts

accurately using the scale on the graph paper I had. When the intercepts are not accurate,

the solution of the linear system will not be accurate.

15. The pentagon has perimeter 58 in.

So, x + y + y + x + 17 = 58 Collect like terms.

2x + 2y = 41

So, one equation is: 2x + 2y = 41

y is greater than x, and the difference between y and x is 1

32

.

So, another equation is: y – x = 1

32

A linear system is:

2x + 2y = 41

y – x = 1

32


Determine the x- and y-intercepts.

For equation :

2x + 2y = 41

Substitute: y = 0 Substitute: x = 0

2x + 2(0) = 41 2(0) + 2y = 41

2x = 41 2y = 41

x = 1

202

y = 1

202

On a grid, plot y as a function of x. Mark a point at 1

202

on each axis. Draw a line through

the points.

For equation :

y – x = 1

32

Substitute: y = 0 Substitute: x = 0

0 – x = 1

32

y – 0 = 1

32

x = –1

32

y = 1

32

A negative value of x makes no sense when we are considering the length of the side of a

pentagon. This value of x is not in the domain. Substitute a different value of x, such as x = 2.



y – 2 = 1

32

y = 1

52

On the grid, mark a point at 1

32

on the y-axis and mark a point at (2, 1

52

). Draw a line

through the points.

From the graph, the coordinates of the point of intersection appear to be (1

82

, 12).

The x-coordinate is approximately 1

82

and the y-coordinate is 12.

So, the value of x is about 1

82

in. and the value of y is about 12 in.


The perimeter, in inches, is: 17 + 21

82

+ 2(12) = 17 + 17 + 24, or 58 in.; this is equal to the

given perimeter.

The difference between y and x is: 12 – 1

82

= 1

32

; this is equal to the given difference.

The solution is correct and the numbers are exact; that is, x is 1

82

in. and y is 12 in.

16. a) 2x + 7y = 3

4x + 3y = 7

To graph this linear system, determine the coordinates of two points on each line.

By looking at the equations, I can tell that the intercepts are fractions, so I choose other

points.

For equation :



Since it is easier to divide by 2 than by 7, I choose values of y to substitute, then solve for

x.

2x + 7y = 3

Substitute: y = 1 Substitute: y = –1

2x + 7(1) = 3 2x + 7(–1) = 3

2x = –4 2x = 10

x = –2 x = 5

Two points on equation have coordinates: (–2, 1) and (5, –1)

For equation :

Since it is easier to divide by 4 than by 3, I choose values of y to substitute, then solve for

x.

4x + 3y = 7

Substitute: y = 1 Substitute: y = –1

4x + 3(1) = 7 4x + 3(–1) = 7

4x = 4 4x = 10

x = 1 x = 2.5

Two points on equation have coordinates: (1, 1) and (2.5, –1)

To plot the decimal values of the coordinates, use a scale of 2 squares to 1 unit on each

axis.

Plot the pair of points for each graph, then draw a line through the points.

From the graph, the solution appears to be: x = 1.8 and y = –0.1

Check whether this solution satisfies both equations.

Substitute x = 1.8 and y = –0.1 in each equation.

For equation :

2x + 7y = 3

L.S. = 2x + 7y R.S. = 3

= 2(1.8) + 7(–0.1)

= 3.6 – 0.7

= 2.9

For equation :

4x + 3y = 7

L.S. = 4x + 3y R.S. = 7

= 4(1.8) + 3(–0.1)

= 7.2 – 0.3

= 6.9

For each equation, the left side is not equal to the right side, but the values are close.

So, the solution x = 1.8 and y = –0.1 is approximate.

b) The solution is approximate because I cannot read the exact fraction or decimal value

from the graph.



C 17. a) Equation has x-intercept 5 and y-intercept 5.

So, the graph of the equation passes through the points with coordinates (5, 0) and (0, 5).

Use the formula for the equation of a line when the coordinates of two points on the line

are known: 1

1

y y

x x

= 2 1

2 1

y y

x x

Substitute: y1 = 0, x1 = 5, y2 = 5, and x2 = 0

0 5 0

5 0 5

y

x

5

y

x = –1 Multiply each side by (x – 5).

y = –1(x – 5)

y = –x + 5 This is equation .

Equation has x-intercept 4 and y-intercept 6.

So, the graph of the equation passes through the points with coordinates (4, 0) and (0, 6).

Use this formula: 1

1

y y

x x

= 2 1

2 1

y y

x x

Substitute: y1 = 0, x1 = 4, y2 = 6, and x2 = 0

0 6 0

4 0 4

y

x

4

y

x =

3

2 Multiply each side by (x – 4).

y = 3

2 (x – 4) Remove brackets.

y = 3

2 x + 6 This is equation .

The linear system is:

y = –x + 5

y = 3

2 x + 6

b) On a grid, plot points at the intercepts for each graph, then draw a line through the points.

The solution appears to be: x = 2 and y = 3


Substitute x = 2 and y = 3 in each equation.

For equation :

y = –x + 5

L.S. = y R.S. = –x + 5



= 3 = –2 + 5

= 3

For equation :

y = 3

2 x + 6

L.S. = y R.S. = 3

2 x + 6

= 3 = 3

2 (2) + 6

= –3 + 6

= 3


18. Graph the equation y = 2x + 1. Substitute two values for x.

When x = 1: When x = –1:

y = 2(1) + 1 y = 2(–1) + 1

y = 3 y = –1

On a grid, plot points at (1, 3) and (–1, –1), then draw a line through them. Extend the line to

the third quadrant. Choose a point on the line in the third quadrant; for example, (–3, –5). Let

this be the coordinates of the point that is the solution of the linear system. To determine a

second equation in the linear system, first choose the coordinates of any other point on the

grid; for example, (1, –3).

Use the formula for the equation of a line when the coordinates of two points on the line are

known: 1

1

y y

x x

= 2 1

2 1

y y

x x

Substitute: y1 = –5, x1 = –3, y2 = –3, and x2 = 1

( 5) 3 ( 5)

( 3) 1 ( 3)

y

x

5

3

y

x =

2

4

5

3

y

x =

1

2 Multiply each side by (x + 3).

y + 5 = 1

2(x + 3)

y + 5 = 1

2x +

3

2



y = 1

2x –

7

2

The second equation could be: y = 1

2x –

7

2

19. a) 2x + 3y = –5

2

x –

3

y = 2

Write each equation in slope-intercept form.

For equation :

2x + 3y = –5 Subtract 2x from each side.

3y = –2x – 5 Divide each side by 3.

y = 2

3 x –

5

3

The slope of the graph of equation is 2

3 .

For equation :

2

x –

3

y = 2 Solve for y. Subtract

2

x from each side.

–3

y = –

2

x + 2 Multiply each side by –3.

–33

y = –3

2

x + (–3)(2) Simplify.

y = 3

2x – 6

The slope of the graph of the equation is 3

2.

The slopes 3

2 and

2

3 are negative reciprocals, so the lines are perpendicular.

b) I chose two numbers that are negative reciprocals; such as 5

3 and

3

5 . I used these

numbers as the slopes of two lines.

I chose two y-intercepts, such as 4 and –2.

To get two equations of a linear system, I substituted a slope and y-intercept into this

form of the equation of a line: y = mx + b

A linear system is:

y = 5

3x + 4

y = 3

5 x – 2

Lesson 7.2 Solving a System of Linear Equations ...hrsbstaff.ednet.ns.ca/mandrecyk/Math...

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Transcript of Lesson 7.2 Solving a System of Linear Equations ...hrsbstaff.ednet.ns.ca/mandrecyk/Math...